Question

高一数学三角函数题——急！！！

已知cos(π/4+x)=4/5,x属于(-π/2,-π/4)，求（sin2x-2sin²x）/（1+tanx）的值

Answer 1

解：cos(π/慧携4+x)=4/5
cos(π/前带伏2+2x)=2cos²(π/4+x)-1=7/25
cos(π/2+2x)=-sin2x
sin2x=-7/25
x∈行烂(-π/2,-π/4)
2x∈(-π,-π/2)
cos2x=-24/25
(sin2x-2sin²x)/(1+tanx)=(sin2x-1+cos2x)/[1+sin2x/(1+cos2x)]
=28/75

Answer 2

cosx -sinx = 4/5*sqrt(2)
sin2x = 1 -(cosx -sinx)^2 = 1 - 32/25 = -7/25;
(cosx+ sinx)^2 = 1+sin2x = 18/25;
cosx+ sinx = - 3sqrt(2)/5;
(sin2x-2sin²x）戚御/（1+tanx） = 2sin(x) (cos(x) -sin(x)/(1+tanx) = sin2x(cosx -sinx)/(cosx+sinx)
= -7/睁搭25*4/5sqrt(2)/高早岩(-3/5sqrt(2) = 28/75

Answer 3

1\2

追问

我要过程，谢谢